# If g(x) = x^4 - 2, find g'(1) and use it to find an equation of the tangent line to the curve y = x^4 - 2 at the point (1,-1) The derivative is 4x^3 so would i be plugging 1 into this, which would...

If g(x) = x^4 - 2, find g'(1) and use it to find an equation of the tangent line to the curve y = x^4 - 2 at the point (1,-1)

The derivative is 4x^3 so would i be plugging 1 into this, which would be the slope at that point in this case 4. so y+1=4(x-1) or y = 4x-5 ????

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To solve for the equation of the line tangent to `y=x^4-2` at (1,-1), take the derivative of y.

`y'=(x^4-2)'`

`y'=(x^4)'-2'`

To get the derivative of the first term, apply the power formula which is `(x^n)'=n*x^(n-1)` .

And for the second term, take note that the derivative of a constant is zero (c'=0).

So,

`y'=4x^(4-1) -0`

`y'=4x^3`

Note that the slope of a line tangent to the curve is equal to the first derivative of the function.

So, to determine the slope of the tangent line substitute the x-coordinate of the point (1, -1).

`y'=4(1)^3`

`y'=4*1`

`y'=4`

Hence, the slope of the tangent line is 4.

Now that the slope is known, apply the point-slope formula to determine the equation of the tangent line.

`y-y_1=m(x-x_1)`

where m is the slope and (x1,y1) is the given point.

Since the given point is (1,-1) and the slope is 4, plug-in x1=1, y1=-1 and m=4.

`y-(-1)=4(x-1)`

`y+1=4x-4`

Then, subtract both sides by 1 to express the equation of the tangent line in the form y=mx+b.

`y+1-1=4x-4-1`

`y=4x-5`

**Hence, the equation of the tangent line is `y=4x-5` .**